Respuesta :
Answer:m
a) 3.11 10²⁴ MJ , b)3.11 10²⁴ MJ, c) 1.03 10³ MJ
Explanation:
Let's look for the potential energy of the satellite based on its height,
F = -dU / dr
dU = - F dr
We integrate and make zero energy for infinite distance
U = G m M int dr / r²
U = G m M (-1 / r)
U = - G m M (1 / r)
With this expression we can calculate the energy that the satellite has for the initial height and subtract it from the energy that the final height, this difference is the energy that we must supply to the system
ΔU = - G m M (1 / rf - 1 / ri)
Let's look for the distance that is from the center of the Earth
r = Re + h
r = 6.37 10⁶ + 210 10³
r = 6.58 10⁶ m
r₀ = 6.37 10⁶ + 99 10³
r₀ = 6,469 10⁶ m
ΔU = 6.67 10⁻¹¹ 994 5.98 10²⁴ (1 / 6.58 10⁶ - 1 / 6.469 10⁶)
ΔU = 3.9647 10¹⁷ (0.152 10⁻⁶ - 0.1546 10⁻⁶)
ΔU = 1.03 10⁹ J
a) the amount of energy to change the orbit satellite is the potential energy plus the kinetic energy
ΔEm = DK + DU
ΔEm = 1.03 10⁹ J + 3.11 10³⁰ J
ΔEm = 3.11 10³⁰ J
ΔEm = 3.11 10²⁴ MJ
Total energy is the sum of the kinetic energy of rotation plus potential energy
b) To calculate the change in kinetic energy, let's use Newton's second law
F = m a
a = v² / r = (wr)² / r = w² r
G m M / r² = m w² r
w² = G M / r³
The change in kinetic energy is
K = ½ I w²
I = m R²
Let's replace
K = ½ (m r²) G m / r³
K = ½ m² G / r
ΔK = ½ m² G (1 / r - 1 / r₀)
ΔK = ½ (5.98 10²⁴)² 6.67 10⁻¹¹ (1 / 6.58 10⁶ m - 1 / 6.469 10⁶)
ΔK = 119.26 10³⁷ (10⁻⁶ 2.61 10⁻³)
ΔK = 311.27 10²⁸ J
ΔK = 3.11 10³⁰ J
ΔK = 3.11 10²⁴ MJ
c) We calculate the change in potential energy
ΔU = 1.03 10⁹ J
ΔU = 1.03 10³ MJ
